Question

Select which of the following suitable substitutes for solving the differential equation \[\dfrac{{dy}}{{dx}} = \left( {4x + y + 1} \right)\].
A \[y = vx\]
B \[y = 4x + v\]
C \[y = 4x\]
D \[y + 4x + 1 = v\]

Answer 1

Hint: Suitable substitution reduces the differential equation in solvable form. To find suitable substitution for a given differential equation, solve the differential equation by substitution method. First use the substitution method then separate the variable. After separation of variables, perform integration to find a general solution.

Formula Used: \[\begin{array}{l}\int {\dfrac{1}{x}dx} = \log x + c\\\int {dx} = x + c\end{array}\]
Here c is an arbitrary constant.

Complete step by step solution: To find suitable substitution for \[\dfrac{{dy}}{{dx}} = \left( {4x + y + 1} \right)\], solve the differential equation.
First multiply by \[dx\]on both sides of the equation.
\[dy = \left( {4x + y + 1} \right)dx\] ……\[\left( 1 \right)\]
It is difficult to separate the variables. So, use a substitution method.
Consider \[4x + y + 1 = v\]and differentiate it.
 \[\begin{array}{l}4x + y + 1 = v\\4dx + dy = dv\end{array}\]
Subtract \[4dx\]from each side of the equation.
\[dy = dv - 4dx\]
Substitute \[4x + y + 1 = v\] and \[dy = dv - 4dx\] in equation \[\left( 1 \right)\].
\[dv - 4dx = vdx\]
Add \[4dx\]on each side of the equation.
\[dv = vdx + 4dx\]
Simplify the equation by taking \[dx\] in common from the right hand side.
\[dv = dx\left( {v + 4} \right)\]
Divide \[v + x\]on each side of the equation.
\[\dfrac{{dv}}{{v + 4}} = dx\]
Integrate both sides of the equation.
 \[\begin{array}{l}\int {\dfrac{{dv}}{{v + 4}}} = \int {dx} \\\log \left( {v + 4} \right) = x + c\end{array}\]
Apply exponential function.
 \[v + 4 = {e^{x + c}}\]
Substitute back \[v = 4x + y + 1\].
\[\begin{array}{l}4x + y + 1 + 4 = {e^{x + c}}\\4x + y + 5 = {e^{x + c}}\end{array}\]
As c is an arbitrary constant, the equation becomes as follows,
\[\begin{array}{l}4x + y + 5 = {e^c}{e^x}\\4x + y + 5 = C{e^x}\end{array}\]
The general solution is \[4x + y + 5 = C{e^x}\].
Hence suitable substitution is \[y + 4x + 1 = v\].

Option ‘D’ is correct

Note: The common mistake students make is selection of wrong substitution. That makes the equation difficult to separate variables.